Question

8 the graph below shows a dilation of triangle 1 to triangle 2 with the origin as the center of dilation. which algebraic rule describes the dilation? a ((x, y)\to(\frac{1}{4} x, \frac{1}{4} y)) b ((x, y)\to(4x, 4y)) c ((x, y)\to(0.4x, 0.4y)) d ((x, y)\to(\frac{1}{2} x, \frac{1}{2} y))

Explanation:

Step1: Identify a vertex from each triangle

Take a vertex from triangle 1 and its corresponding vertex in triangle 2. For example, in triangle 1, let's take the vertex at (1, 2), and in triangle 2, the corresponding vertex is at (4, 8).

Step2: Determine the scale factor

To find the scale factor \( k \), we divide the coordinates of the image (triangle 2) by the coordinates of the pre - image (triangle 1). For the x - coordinate: \( \frac{4}{1}=4 \), for the y - coordinate: \( \frac{8}{4} = 2\)? Wait, no, let's take another vertex. Let's take the vertex of triangle 1 at (1,1) and the corresponding vertex in triangle 2 at (4,4). Then for x - coordinate: \( \frac{4}{1}=4 \), for y - coordinate: \( \frac{4}{1}=4 \). Another vertex: triangle 1 at (2,1), triangle 2 at (6,4)? Wait, no, looking at the graph, triangle 1 has vertices at (1,1), (1,2), (2,1) and triangle 2 has vertices at (4,4), (4,8), (6,4). So for the vertex (1,1) in triangle 1, the corresponding vertex in triangle 2 is (4,4). So the transformation is \( (x,y)\to(4x,4y) \) because when \( x = 1,y = 1 \), \( 4x=4 \), \( 4y = 4 \); when \( x=1,y = 2 \), \( 4x = 4 \), \( 4y=8 \); when \( x = 2,y=1 \), \( 4x=8 \)? Wait, no, the third vertex of triangle 2 is (6,4). Wait, maybe I misread the vertices. Let's re - examine: Triangle 1: points at (1,1), (1,2), (2,1). Triangle 2: points at (4,4), (4,8), (6,4). So for (1,1) to (4,4): \( 4=4\times1 \), \( 4 = 4\times1 \); (1,2) to (4,8): \( 4=4\times1 \), \( 8=4\times2 \); (2,1) to (6,4): Wait, \( 6 = 3\times2 \), \( 4=4\times1 \). Wait, maybe I made a mistake. Wait, the center of dilation is the origin. Wait, no, the center of dilation is the origin? Wait, the problem says "the origin as the center of dilation", but the triangles are not centered at the origin. Wait, maybe the vertices of triangle 1 are (1,1), (1,2), (2,1) and triangle 2 are (4,4), (4,8), (6,4). Let's check the scale factor. The distance from the center of dilation (origin) to (1,1) is \( \sqrt{1^2 + 1^2}=\sqrt{2} \), and to (4,4) is \( \sqrt{4^2+4^2}=\sqrt{32}=4\sqrt{2} \). The ratio of distances is \( \frac{4\sqrt{2}}{\sqrt{2}} = 4 \). Similarly, distance from origin to (1,2) is \( \sqrt{1 + 4}=\sqrt{5} \), distance to (4,8) is \( \sqrt{16 + 64}=\sqrt{80}=4\sqrt{5} \), ratio is 4. Distance from origin to (2,1) is \( \sqrt{4 + 1}=\sqrt{5} \), distance to (6,4) is \( \sqrt{36+16}=\sqrt{52}=2\sqrt{13} \)? Wait, no, maybe the center of dilation is not the origin? Wait, the problem says "the origin as the center of dilation". Wait, maybe my vertex identification is wrong. Let's look at the coordinates again. The grid has x from 1 to 10 and y from 1 to 10. Triangle 1: at x = 1, y = 1; x = 1, y = 2; x = 2, y = 1. Triangle 2: at x = 4, y = 4; x = 4, y = 8; x = 6, y = 4. So the vector from the center of dilation (origin) to (1,1) is \( \vec{v_1}=(1,1) \), and to (4,4) is \( \vec{v_2}=(4,4)=4\times(1,1) \). The vector from origin to (1,2) is \( (1,2) \), and to (4,8) is \( (4,8)=4\times(1,2) \). The vector from origin to (2,1) is \( (2,1) \), and to (6,4) is \( (6,4)=3\times(2, \frac{4}{3}) \). Wait, that's not 4 times. Wait, maybe the center of dilation is not the origin? But the problem says it is. Wait, maybe I misread the vertices. Let's check the y - axis: triangle 1 has a vertex at y = 2, triangle 2 at y = 8. 8/2 = 4. x - axis: triangle 1 has a vertex at x = 1, triangle 2 at x = 4. 4/1=4. Another vertex: triangle 1 at x = 2, triangle 2 at x = 6? No, 6/2 = 3. Wait, no, maybe the vertices of triangle 1 are (1,1), (1,2), (2,1) and triangle 2 are (4,4), (4,8), (6,4). So the transformation from (1,1) to (4,4) is (x,y)→(4x,4y) because 1×4 = 4, 1×4 = 4. From (1,2) to (4,8): 1×4 = 4, 2×4 = 8. From (2,1) to (6,4): Wait, 2×3=6, 1×4 = 4. That doesn't match. Wait, maybe the third vertex of triangle 1 is (2,1) and the third vertex of triangle 2 is (6,4). So 2×3=6, 1×4 = 4. That's inconsistent. Wait, maybe I made a mistake in the vertex selection. Let's look at the vertical side: triangle 1 has a vertical side from (1,1) to (1,2), length 1. Triangle 2 has a vertical side from (4,4) to (4,8), length 4. The horizontal side: triangle 1 has a horizontal side from (1,1) to (2,1), length 1. Triangle 2 has a horizontal side from (4,4) to (6,4), length 2? Wait, no, (4,4) to (6,4) is length 2, (1,1) to (2,1) is length 1. So the scale factor for x - direction is 2? No, 4 - 6 is 2 units, 1 - 2 is 1 unit. Wait, the vertical side: 8 - 4 = 4 units, 2 - 1 = 1 unit. So scale factor 4 for y - direction, 2 for x - direction? No, that can't be. Wait, the problem says "dilation", which is a similarity transformation, so the scale factor should be the same for x and y. So maybe the vertices of triangle 1 are (1,1), (1,2), (2,1) and triangle 2 are (4,4), (4,8), (8,4). Then (2,1) to (8,4): 2×4 = 8, 1×4 = 4. Yes! So I misread the third vertex of triangle 2. It should be (8,4) instead of (6,4). So triangle 2 has vertices (4,4), (4,8), (8,4). Then: (1,1)→(4,4): 1×4 = 4, 1×4 = 4; (1,2)→(4,8): 1×4 = 4, 2×4 = 8; (2,1)→(8,4): 2×4 = 8, 1×4 = 4. Now it's consistent. So the transformation is \( (x,y)\to(4x,4y) \), which is option B.

Answer:

B. \((x, y)\to(4x, 4y)\)